The complete proofs are available in my Idris github repo, and if you want to get the most out of the code, it's probably best to step through the types in Idris. In this post I'll talk through a few of the details.

Directly proving the reverse1 (reverse1 x) = x by induction is hard, since the function doesn't really follow so directly. Instead we need to define a helper lemma.

Coming up with these helper lemmas is the magic of writing proofs - and is part intuition, part thinking about what might be useful and part thinking about what invariants are obeyed. With that lemma, you can prove by induction on the first argument (which is the obvious one to chose since ++ evaluates the first argument first). Proving the base case is trivial:

The proof immediately reduces to reverse1 ys == reverse1 ys ++ [] and we can invoke the standard proof that if ++ has [] on the right we can ignore it. After applying that rewrite, we're straight back to Refl.

The :: is not really any harder, we immediately get to reverse1 ys ++ (reverse1 xs ++ [x]), but bracketed the wrong way round, so have to apply appendAssociative to rebracket so it fits the inductive lemma. The proof looks like:

For the remaining two proofs, I first decided to prove reverse1 x = reverse2 x and then prove the reverse2 (reverse2 x) = x in terms of that. To prove equivalence of the two functions I first needed information about the fundamental property that rev2 obeys:

Namely that the accumulator can be tacked on the end. The proof itself isn't very complicated, although it does require two calls to the inductive hypothesis (you basically expand out rev2 on both sides and show they are equivalent):

In essence the proof_rev term shows that rev behaves like the O(n^2) reverse.

Finally, actually proving that reverse2 (reverse2 x) is true just involves replacing all the occurrences of reverse2 with reverse1, then applying the proof that the property holds for reverse1. Nothing complicated at all: